To solve this problem it is necessary to make a correlation between the energy lost by the frictional force and the kinetic energy by the puck.
Considering the definitions for the friction force as a product between the coefficient of friction and the normal force (mass by gravity) we have to:
[tex]W_f = KE[/tex]
[tex]F_f \Delta x = \frac{1}{2}mv^2[/tex]
[tex](\mu mg)\Delta x = \frac{1}{2} mv^2[/tex]
Where,
[tex]\Delta x[/tex] = Distance traveled
v = Velocity of the puck
m = mass of the puck
g = Gravitational acceleration
[tex]\mu =[/tex] Coefficient of kinetic friction
Re-arrange to find the Coefficient of kinetic friction
[tex](\mu mg)\Delta x = \frac{1}{2} mv^2[/tex]
[tex](\mu g)\Delta x = \frac{1}{2} v^2[/tex]
[tex]\mu = \frac{1}{2} \frac{v^2}{g\Delta x}[/tex]
Replacing we have finally,
[tex]\mu = \frac{1}{2} \frac{4.4^2}{(9.8)(22)}[/tex]
[tex]\mu = 0.0448[/tex]
Therefore the coefficient of kinetic friction between the ice and the puck is 0.0448
